Question

The radius of \(Na^+\) is \(95\,pm\) and that of \(Cl^-\) ion is \(181\,pm\). Hence, the coordination number of \(Na^+\) will be

• A. 4
• B. 6
• C. 8
• D. unpredictable

Hint:

Use radius ratio \(r_+/r_-\). If it lies between \(0.414\) and \(0.732\), the coordination number is 6 (octahedral).

Answer 1

The Correct Option is B

Step 1: Use radius ratio rule.
\[ \frac{r_+}{r_-} = \frac{95}{181} \approx 0.525 \] Step 2: Compare with standard limits.
Radius ratio ranges:
- CN = 4 (tetrahedral): \(0.225 - 0.414\)
- CN = 6 (octahedral): \(0.414 - 0.732\)
- CN = 8 (cubic): \(0.732 - 1.0\)
Step 3: Locate 0.525.
Since
\[ 0.414<0.525<0.732 \] coordination number is 6.
Final Answer: \[ \boxed{6} \]